Answer
$$\frac{1}{5}{\left( {{x^2} + 1} \right)^{5/2}} - \frac{1}{3}{\left( {{x^2} + 1} \right)^{3/2}} + C $$
Work Step by Step
$$\eqalign{
& \int {{x^3}\sqrt {{x^2} + 1} } dx \cr
& = \int {{x^2}\sqrt {{x^2} + 1} } xdx \cr
& {\text{Integrate by the substitution mehtod}} \cr
& \,\,\,\,\,{\text{Let }}u = {x^2} + 1,\,\,\,\,{x^2} = u - 1,\,\,\,\,\,du = 2xdx, \cr
& \int {{x^2}\sqrt {{x^2} + 1} } xdx = \int {\left( {u - 1} \right)\sqrt u } \left( {\frac{1}{2}du} \right) \cr
& = \frac{1}{2}\int {\left( {u - 1} \right){u^{1/2}}} du \cr
& = \frac{1}{2}\int {\left( {{u^{3/2}} - {u^{1/2}}} \right)} du \cr
& \cr
& {\text{Then}}{\text{,}} \cr
& = \frac{1}{2}\left( {\frac{{{u^{5/2}}}}{{5/2}} - \frac{{{u^{3/2}}}}{{3/2}}} \right) + C \cr
& = \frac{1}{2}\left( {\frac{2}{5}{u^{5/2}} - \frac{2}{3}{u^{3/2}}} \right) + C \cr
& = \frac{1}{5}{u^{5/2}} - \frac{1}{3}{u^{3/2}} + C \cr
& {\text{write in terms of }}x{\text{; substitute }}{x^2} + 1{\text{ for }}u \cr
& = \frac{1}{5}{\left( {{x^2} + 1} \right)^{5/2}} - \frac{1}{3}{\left( {{x^2} + 1} \right)^{3/2}} + C \cr} $$
